Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 15, Problem 104AP

The equilibrium constant K p for the following reaction is 4 .31  ×  10 -4  at 375°C:

N 2 (g)+3H 2 ( g ) 2 NH 3 ( g )

In a certain experiment a student starts with 0.862 atm of N2 and 0.373 atm of H2 in a constant-volume vessel at 375°C . Calculate the partial pressures of all species when equilibrium is reached.

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The partial pressure of the species present in the given equilibrium reaction is to be calculated with given KP value.

Concept introduction:

Equilibrium constant for gas phase reactions is written in terms of the partial pressures of the species present in the reactionas the concentration of gases is directly proportional to the partial pressures. The equilibrium constant in terms of partial pressure is denoted by KP and whether the reactants are present in the solid or liquid phase, is not represented in the KP expression.

Equilibrium constant is the ratio of the concentration of reactants and products present in the chemical reaction.

For a general reaction: aA+bBcC+dD

The general formula for writing equilibrium expression for the reaction is given as:

KC=[C]ceqm[D]deqm[A]aeqm[B]beqm

Here, KC is the equilibrium constant and C in KC stands for the concentration. [A], [B], [C], and [D] are the equilibrium concentration of reactants A and B and product C and D, respectively.

A and B are reactants, C is products, and x,y, and z are their respective stoichiometric coefficients.

For the general reaction: aA+bBcC+dD

The general formula for writing equilibrium expression for the reaction is given as:

Kp=PCcPDdPAaPBb

Here, p in Kp stands for the pressure. PA,PB,PC,andPD are the equilibrium partial pressure of reactants A and B and products C and D.

A and B are reactants, C and D are products, and a,b,c, and d are their respective stoichiometric coefficients.

Equilibrium constants of gas phase reaction are written in terms of partial pressures because concentration of gases is directly proportional to partial pressures.

The relationship between KP and KC is given as:

KP=KC(RT)ΔnKc=KP(RT)Δn

Answer to Problem 104AP

Solution: PN2=0.860atm,PH2=0.366atm, and PNH3=4.40×103atm.

Explanation of Solution

Given information: Kp=4.31×104

The given balanced equation is as follows:

N2(g)+H2(g)2NH3(g)

The initial change equilibrium table for the reaction is as follows:

N2H2NH3Initial0.8620.3730Changex3x+2xEquilibrium(0.862x)(0.3733x)2x

The equilibrium constant expression for the reaction is given as follows:

Kp=(PNH3)2(PN2)(PH2)3

Here, Kp is the equilibrium constant of pressure, PN2 is the partial pressure of N2, PH2 is the partial pressure of H2, and PNH3 is the partial pressure of NH3.

Substitute the values of Kp, PN2, PH2, and PNH3 in the above equation,

4.31×104=(2x)2(0.862x)(0.3733x)3

As the value of x is too small, therefore, the expression (0.862x) is nearly equal to 0.862 and the expression (0.3733x) is nearly equal to 0.373.

4.31×104=(2x)2(0.862)(0.373)3x=4.82×106x=2.20×103

The partial pressure of N2 is calculated as follows:

PN2=(0.8622.20×103)atm=0.860atm

The partial pressure of H2 is calculated as:

PN2=(0.3733×2.20×103)atm=0.366atm

The partial pressure of NH3 is calculated as follows:

PNH3=2×2.20×103atm=4.40×103atm

Conclusion

The partial pressures of N2, H2, and NH3 in the given reaction are 0.860atm, 0.366atm, and 4.40×103atm, respectively.

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Chapter 15 Solutions

Chemistry

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